PhysicsBowl 2011 – Ivy Learner (2024)

Which one of the following choices best represents the average mass of an adult human male?

$\textbf{(A) }$ $1.0\times10^{-2}\ \text{kg} \qquad\newline$ $\textbf{(B) }$ $1.0\times10^{-1}\ \text{kg} \qquad\newline$ $\textbf{(C) }$ $1.0\times10^{1}\ \text{kg} \qquad\newline$ $\textbf{(D) }$ $1.0\times10^{2}\ \text{kg} \qquad\newline$ $\textbf{(E) }$ $1.0\times10^{3}\ \text{kg}$

$\textbf{D}$

The average human male is closest to 100 kg. I am glad to see this is a question about the weight of a male but not female...

Albert Einstein’s most famous equation is $E=mc^2$. The unit for the quantity represented by $E$ can be written as which one of the following options?

$\textbf{(A) }$ seconds$ \qquad$ $\textbf{(B) }$ newtons$ \qquad$ $\textbf{(C) }$ kilograms$ \qquad$ $\textbf{(D) }$ meters$ \qquad$ $\textbf{(E) }$ joules

$\textbf{E}$

The unit of energy (E) is joules.

The distance of one thousand meters can be written as one kilometer. Which one of the following choices best represents a distance of one million meters?

$\textbf{(A) }$ one gigameter$ \qquad\newline$ $\textbf{(B) }$ one megameter $ \qquad\newline$ $\textbf{(C) }$ one micrometer$ \qquad\newline$ $\textbf{(D) }$ one nanometer $ \qquad\newline$ $\textbf{(E) }$ one terameter

$\textbf{B}$

The metric prefix for million (or $10^6$) is mega.

The following two measurements of length are recorded from an experiment: $L_1=4.57\times10^3\ \text{m}$ and $L_2=2.130\times10^2\ \text{m}$. Using the rules of significant digits, what is the sum $(L_1+L_2)$ of these measurements?

$\textbf{(A) }$ $5\times10^3\ \text{m} \qquad$ $\textbf{(B) }$ $4.8\times10^3\ \text{m} \qquad$ $\textbf{(C) }$ $4.78\times10^3\ \text{m} \qquad$ $\textbf{(D) }$ $4.783\times10^3\ \text{m} \qquad$ $\textbf{(E) }$ $6.70\times10^3\ \text{m}$

$\textbf{C}$

Rules of addition look at the precision of a measurement. In other words, we look at the last column for which each value has a number… lining up these values gives a sum of $(4.57+0.213)\times10^3\ \text{m}$. Since the first value only has two places past the decimal (and the other value has 3 places), the sum must end 2 places past the decimal. Hence, $(4.57+0.213)\times10^3\ \text{m}=4.78\times10^3\ \text{m}$

A positively charged rod is brought near a metal electroscope that is initially uncharged. As shown in the figure, the rod does not touch the electroscope. There is no charge transfer between the electroscope and rod, but the leaves of the electroscope move apart from each other when the rod is brought near the top of the electroscope. Which one of the following choices is the best explanation for this phenomenon?

$\textbf{B}$

Protons in the metal are not free to move as they are in the nuclei of the atoms. The mobile charges here are electrons.

When the ball reaches its maximum height, which one of the following choices best represents the magnitude of the net force acting on the ball?

$\textbf{(A) }$ 0 N$ \qquad$ $\textbf{(B) }$ 0.100 N$ \qquad$ $\textbf{(C) }$ 1.00 N$ \qquad$ $\textbf{(D) }$ 10.0 N$ \qquad$ $\textbf{(E) }$ 12.0 N

$\textbf{C}$

The net force acting on the ball is the gravitational force $G=mg=(0.100)(10)=1.00\ \text{N}$.

Approximately how much time does it take the ball to reach the maximum height from its launch?

$\textbf{(A) }$ 0.65 s$ \qquad$ $\textbf{(B) }$ 1.10 s$ \qquad$ $\textbf{(C) }$ 1.20 s$ \qquad$ $\textbf{(D) }$ 1.55 s$ \qquad$ $\textbf{(E) }$ 2.40 s

$\textbf{D}$

The time for the ball to reach the top is the same as the time of free fall, which is $t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2(12)}{10}}=1.55\ \text{s}$.

$\textbf{E}$

In order for the speed of the object to increase, the velocity and acceleration must have the same sign. This means that the acceleration is directed with the velocity, thereby increasing the size of the velocity vector (the speed).

hich one of the following choices best represents the speed of 60.0 miles/hr rewritten in units of km/day?

$\textbf{(A) }$ 2300$ \qquad$ $\textbf{(B) }$ 1440$ \qquad$ $\textbf{(C) }$ 900$ \qquad$ $\textbf{(D) }$ 4.00$ \qquad$ $\textbf{(E) }$ 1.56

$\textbf{A}$

$60.0\ \dfrac{\text{miles}}{\text{hr}}=60.0\ \dfrac{1.609\ \text{km}}{\frac{1}{24}\text{day}}=2320\ \dfrac{\text{km}}{\text{day}}$

A string of negligible mass connects an object of mass $M=10\ \text{kg}$ to the ceiling of an elevator. The elevator experiences a constant downward speed of 3.0 m/s. Let $T$ represent the magnitude of the force on the mass $M$ by the string (tension), $G$ represent the magnitude of the gravitational force by the Earth acting on the mass $M$ hanging in the elevator, and $F$ represent the magnitude of the net force acting on the mass $M$. Which one of the following choices describes the relationships between these forces?

$\textbf{E}$

Since the object is moving with a constant velocity, its acceleration must be zero, thus the net force $F=0$. So the tension and the gravitational force acting on the object are equal. Hence we get $F<T=G$.

The largest particle accelerator in the world is called the LHC. What does LHC stand for?

$\textbf{(A) }$ Large Hadron Conductor$ \qquad\newline$
$\textbf{(B) }$ Light Hadron Container$ \qquad\newline$
$\textbf{(C) }$ Light Hadron Collider$ \qquad\newline$
$\textbf{(D) }$ Large Hadron Container$ \qquad\newline$
$\textbf{(E) }$ Large Hadron Collider

$\textbf{E}$

The LHC is the Large Hadron Collider.

A 5.0 kg mass moves along the $x-\text{axis}$. At one instant of time, the mass has position 2.0 m, velocity 3.0 m/s, and acceleration 4.0 m/s$^2$. What is the net force acting on the mass at this instant of time?

$\textbf{(A) }$ 0 N$ \qquad$ $\textbf{(B) }$ 10.0 N$ \qquad$ $\textbf{(C) }$ 15.0 N$ \qquad$ $\textbf{(D) }$ 20.0 N$ \qquad$ $\textbf{(E) }$ 22.5 N

$\textbf{D}$

By Newton’s Second Law, the net force acting on the mass is $F=ma=(5)(4)=20.0\ \text{N}$.

A physics book sits at rest on a table. On top of the book is a calculator, also at rest. Which one of the following choices is the Newton’s Third Law pair force to the force that the table exerts on the book?

$\textbf{A}$

When the table exerts a force on the book, by Newton’s Third Law, the book also exerts a force on the table.

A car makes a trip in two parts:$\newline$
Part 1: it travels a distance of 800 m at a constant speed of 4.0 m/s.$\newline$
Part 2: it travels a distance of 1200 m at a constant speed of 20.0 m/s.$\newline$
What is the average speed of the car for the entire two-part trip?

$\textbf{(A) }$ 7.7 m/s$ \qquad$ $\textbf{(B) }$ 10.4 m/s$ \qquad$ $\textbf{(C) }$ 12.0 m/s$ \qquad$ $\textbf{(D) }$ 13.6 m/s$ \qquad$ $\textbf{(E) }$ 17.3 m/s

$\textbf{A}$

The total distance the car traveled is $800+1200=2000\ \text{m}$, and it takes $\dfrac{800}{4}+\dfrac{1200}{20}=260\ \text{s}$. So the average speed is $\dfrac{2000}{260}=7.7\ \text{m/s}$.

A 4.0 kg object is pushed to the right on a rough surface by a horizontal force of 15 N as shown. During the push, the object accelerates uniformly to the right at $2.50\ \text{m/s}^2$. What must be the magnitude of the force of friction also acting on the object during the push?

$\textbf{B}$

By Newton’s Second Law, The net force acting on the object is $F-f=ma$. So the frictional force $f=F-ma=15-(4)(2.5)=5\ \text{N}$.

Which one of the following is NOT a vector quantity?

$\textbf{(A) }$ Linear momentum$ \qquad\newline$
$\textbf{(B) }$ Electric field$ \qquad\newline$
$\textbf{(C) }$ Average velocity$ \qquad\newline$
$\textbf{(D) }$ Instantaneous acceleration$ \qquad\newline$
$\textbf{(E) }$ Kinetic energy

$\textbf{E}$

Energy is a scalar quantity and has no direction associated with it.

A new radio station broadcasts its signal with a wavelength of 3.40 m. At what frequency is this station broadcasting?

$\textbf{(A) }$ $1.76\times10^8\ \text{Hz} \qquad$ $\textbf{(B) }$ $8.82\times10^7\ \text{Hz} \qquad$ $\textbf{(C) }$ $100\ \text{Hz} \qquad$ $\textbf{(D) }$ $0.294\ \text{Hz} \qquad$ $\textbf{(E) }$ $1.13\times10^{-8}\ \text{Hz}$

$\textbf{B}$

The frequency $f=\dfrac{c}{\lambda}=\dfrac{3\times10^8}{3.40}=8.82\times10^7\ \text{Hz}$.

Which one of the following choices best represents the average acceleration of the object during the time interval from $t=4.0\ \text{s}$ to $t=9.0\ \text{s}$?

$\textbf{(A) }$ $0.80\ \text{m/s}^2 \qquad$ $\textbf{(B) }$ $0\ \text{m/s}^2 \qquad$ $\textbf{(C) }$ $-0.80\ \text{m/s}^2 \qquad$ $\textbf{(D) }$ $-1.6\ \text{m/s}^2 \qquad$ $\textbf{(E) }$ $-3.2\ \text{m/s}^2$

$\textbf{D}$

By definition, the average acceleration is found as the change in the velocity divided by the time. Hence, we compute from the points on the graph that $\bar{a}=\dfrac{\Delta v}{\Delta t}=\dfrac{(-4)-(4)}{5}=-1.6\ \text{m/s}^2$.

Which one of the following choices best represents the instantaneous acceleration of the object at the time of $t=5.0\ \text{s}$?

$\textbf{(A) }$ $0\ \text{m/s}^2 \qquad$ $\textbf{(B) }$ $-1.6\ \text{m/s}^2 \qquad$ $\textbf{(C) }$ $-2.0\ \text{m/s}^2 \qquad$ $\textbf{(D) }$ $-3.2\ \text{m/s}^2 \qquad$ $\textbf{(E) }$ $-4.0\ \text{m/s}^2$

$\textbf{E}$

The line segment at $t=5.0\ \text{s}$ looks like a straight line passing (4,4) and (6,-4). The acceleration is the slope of the tangent, which is the straight line itself. Thus the acceleration is $\dfrac{4-(-4)}{4-6}=-4.0\ \text{m/s}^2$.

The transistor is a solid piece of semiconducting material with many applications in electronics. The initial development of transistors and the Nobel Prize awarded for its invention occurred closest to which calendar year?

$\textbf{(A) }$ 1890$ \qquad$ $\textbf{(B) }$ 1920$ \qquad$ $\textbf{(C) }$ 1950$ \qquad$ $\textbf{(D) }$ 1980$ \qquad$ $\textbf{(E) }$ 2005

$\textbf{C}$

The transistor was introduced around 1947 with the Nobel Prize awarded for its invention in 1956.

Which one of the following choices best represents the time it takes for light coming from the Moon’s surface to reach the Earth?

$\textbf{(A) }$ 0.0001 s$ \qquad$ $\textbf{(B) }$ 0.10 s$ \qquad$ $\textbf{(C) }$ 1.0 s$ \qquad$ $\textbf{(D) }$ 10.0 s$ \qquad$ $\textbf{(E) }$ 1 minute

$\textbf{C}$

The key to the problem is to find the distance $r$ between the earth and the moon. Using Kepler’s Third Law, $\dfrac{r^3}{T^2}=\dfrac{GM_e}{4\pi^2}$. The period $T$ of the moon is about 1 month. Hence we can find the distance $r=\sqrt[3]{\dfrac{GM_eT^2}{4\pi^2}}=\sqrt[3]{\dfrac{(6.7\times10^{-11})(6.0\times10^{24})(30\times24\times60\times60)^2}{4\pi^2}}=4.1\times10^8\ \text{m}$. The time for the light to travel is $t=\dfrac{r}{c}=\dfrac{4.1\times10^8}{3\times10^8}=1.4\ \text{s}$.

Three moles of an ideal gas occupy a container of volume $2.5\times10^{-2}\ \text{m}^3$. The pressure inside this container is $5.00\times10^5\ \text{Pa}$. Which one of the following choices best represents the temperature of the gas?

$\textbf{(A) }$ 225 K$ \qquad$ $\textbf{(B) }$ 500 K$ \qquad$ $\textbf{(C) }$ 775 K$ \qquad$ $\textbf{(D) }$ 1500 K$ \qquad$ $\textbf{(E) }$ 50800 K

$\textbf{B}$

According to the ideal gas equation, the temperature $T=\dfrac{PV}{nR}=\dfrac{(5\times10^5)(2.5\times10^{-2})}{(3)(8.31)}=501\ \text{K}$.

$\textbf{A}$

Wave speed $v$ for a string depends on the tension $T$ in the string and the mass per unit length $\mu$ as $v=\sqrt{\dfrac{T}{\mu}}$. Since these have not changed, the wave speed is unchanged. Using $v=\lambda f$, an increase in the frequency means that the wavelength decreases. So, doubling the frequency results in halving the wavelength.

The product (3 Teslas)×(2 meters)×(4 meters/second) is equivalent to which one of the following?

$\textbf{(A) }$ 24 Amperes$ \qquad$ $\textbf{(B) }$ 24 Coulombs $ \qquad$ $\textbf{(C) }$ 24 Watts $ \qquad$ $\textbf{(D) }$ 24 Ohms$ \qquad$ $\textbf{(E) }$ 24 Volts

$\textbf{E}$

The product shown is of magnetic field by length by speed… in symbols, this is $Blv$ which can represent an emf which has units of volts.

During the New Moon phase of the lunar cycle, approximately what percent of the Moon’s surface receives light from the Sun?

$\textbf{(A) }$ $0\ \% \qquad$ $\textbf{(B) }$ $25.0\ \%\qquad$ $\textbf{(C) }$ $50.0\ \% \qquad$ $\textbf{(D) }$ $ 75.0\ \%\qquad$ $\textbf{(E) }$ $100\ \%$

$\textbf{C}$

The Moon always receives sunlight on approximately $50\%$ of its surface. What we see is a function of the location of the Moon in its orbit and whether the illuminated side of the Moon is facing the Earth.

The density of a solid cubic block is given as $\rho=0.750\ \text{g/cm}^3$. This block is dropped into a large swimming pool filled with water. When equilibrium is established and all motion stops, which choice best describes the location of the block? The block is very small in comparison to the pool.

$\textbf{(A) }$ The block is resting at the bottom of the pool.$ \qquad\newline$
$\textbf{(B) }$ The block is completely submerged, but stops $75\ \%$ of the way to the bottom of the pool.$ \qquad\newline$
$\textbf{(C) }$ The block is completely submerged, but stops $25\ \%$ of the way to the bottom of the pool.$ \qquad\newline$
$\textbf{(D) }$ The block is floating with $75\ \%$ of its volume under the water’s surface.$ \qquad\newline$
$\textbf{(E) }$ The block is floating with $25\ \%$ of its volume under the water’s surface.

$\textbf{D}$

Since the density of the material is less than the density of water, the object will float. The object must displace enough water to create a buoyant force to balance the gravitational force. So, from the free body diagram analysis, we have $\rho_{water}gV_{underwater}=\rho_{block}gV_{total}$. Solving for the fraction below the water, we compute $\dfrac{V_{underwater}}{V_{total}}=\dfrac{\rho_{block}}{\rho_{water}}=75\ \%$.

A person throws an object of mass $M$ straight upward with an initial non-zero speed $v$. The object rises to a maximum height $H$ above the launch point. The person now throws an object of mass $\dfrac12M$ straight upward with initial speed $2v$. In terms of $H$, to what maximum height does the object of mass $\dfrac12M$ rise above the launch point? Ignore air resistance.

$\textbf{(A) }$ $8H \qquad$ $\textbf{(B) }$ $4H \qquad$ $\textbf{(C) }$ $2H \qquad$ $\textbf{(D) }$ $\sqrt2H \qquad$ $\textbf{(E) }$ $H$

$\textbf{B}$

The height $H=\dfrac{v^2}{2g}$. For the second object with twice the speed, the height will be $4H$.

An electron moves with speed $1.00\times10^3\ \text{m/s}$ as it enters a region of space that has only a uniform magnetic field. The electron is accelerated because of this field with a constant magnitude of $8.00\times10^5\ \text{m/s}^2$ for the entire time of $1.00\times10^{-3}\ \text{s}$ that it is in the field. With what speed would the electron exit the field? Ignore gravity.

$\textbf{(A) }$ $1.80\times10^3\ \text{m/s} \qquad$ $\textbf{(B) }$ $1.41\times10^3\ \text{m/s} \qquad$ $\textbf{(C) }$ $1.00\times10^3\ \text{m/s} \qquad$ $\textbf{(D) }$ $2.00\times10^2\ \text{m/s} \qquad\newline$
$\textbf{(E) }$ It depends on the initial angle of the electron’s velocity with respect to the magnetic field.

$\textbf{C}$

The magnetic force does no work on the charged particle (it is a deflecting force resulting in a change in direction of travel). Consequently, the speed of the particle never changes while in the magnetic field, independent of its direction with respect to the field.

Two cars travel to the right, each starting from rest, along a straight road. Car A has twice the acceleration of Car B. After traveling a distance $d$, Car A has speed $v$. When Car B has traveled the same distance $d$, what is its speed in terms of $v$?

$\textbf{(A) }$ $\dfrac14v \qquad$ $\textbf{(B) }$ $\dfrac12v \qquad$ $\textbf{(C) }$ $\dfrac{\sqrt3}{2}v \qquad$ $\textbf{(D) }$ $\dfrac{\sqrt2}{2}v \qquad$ $\textbf{(E) }$ $v$

$\textbf{D}$

By the equation $v^2=2ax$, when halving the acceleration, the speed will be $\dfrac{\sqrt2}{2}v$.

Approximately how much energy is required to transform a 20.0 g cube of ice at a temperature of $-10.0^\circ\text{C}$ into liquid water at a temperature of $10.0^\circ\text{C}$?

$\textbf{(A) }$ 840 J$ \qquad$ $\textbf{(B) }$ 1300 J$ \qquad$ $\textbf{(C) }$ 1700 J$ \qquad$ $\textbf{(D) }$ 7900 J$ \qquad$ $\textbf{(E) }$ 46000 J

$\textbf{D}$

Adding energy to the ice results in the ice changing its temperature to zero degrees, melting into water, and then having its temperature rise to 10 degrees. The computation of this energy is done as $$Q=mc_{ice}\Delta T+mL_f+mc_{water}\Delta T$$ As a result, $$Q=(0.020\ \text{kg})\left[(2100\ \text{J/kg}\cdot\text{K})(10\ \text{K})+(3.3\times10^5\ \text{J/kg})+(4200\ \text{J/kg}\cdot\text{K})(10\ \text{K})\right]=7900\ \text{J}$$

An ideal green pigment is mixed with an ideal red pigment. After mixing, which one of the following choices best represents the color of the pigment produced?

$\textbf{(A) }$ Blue$ \qquad$ $\textbf{(B) }$ Cyan$ \qquad$ $\textbf{(C) }$ Magenta$ \qquad$ $\textbf{(D) }$ Yellow$ \qquad$ $\textbf{(E) }$ Black

$\textbf{E}$

Pigments work on the subtractive color system. This means that a blue pigment is blue because it reflects blue light and absorbs both red and green light. The green pigment absorbs the red and blue, reflecting only green. When mixed, all three primary colors will be absorbed, meaning that none of the colors are reflected and the pigment is very dark, making (E) (black) the correct response.

For the set-up shown of a candle and screen placed 30 cm apart, there are two locations at which a thin converging lens can be placed to produce a focused real image. One real image of the candle appears on the screen when the lens is located 12 cm from the candle. From this location, how does this lens now have to be moved in order to make the second real image of the candle appear on the screen?

$\textbf{C}$

When the lens is placed 12 cm away from the candle, the object distance is $d_o=12\ \text{cm}$, and the image distance is $d_i=30-12=18\ \text{cm}$. The lens equation is $\dfrac1{d_o}+\dfrac1{d_i}=\dfrac1f$. Apparently, by exchanging the value of $d_o$ and $d_i$, the equation remains solid. Thus we know the lens can be move toward the screed by $18-12=6\ \text{cm}$.

A wire lies in the plane of the page with conventional current $I$ shown. At the instant shown, what is the direction of the magnetic force on the electron that is moving directly to the right toward the wire?

$\textbf{B}$

By the right-hand rule, with the right thumb along the current, the right fingers wrap in the sense of the magnetic field from the wire. This means that the magnetic field from the wire is out of the plane of the page at the location of the electron. To find the force on the electron, now we point the right fingers in the direction of the velocity and curl them into the direction of the magnetic field. The thumb points in the direction of the force (toward the bottom of the page) EXCEPT that since the charge in question is negative, the hand has to be turned 180 degrees which results in the right thumb pointing toward the top of the page and in the direction of the force.

Mary rides her bicycle directed due South. She applies the brakes quickly bringing the bicycle to rest with the wheels rolling without slipping on the ground. Which one of the following choices best represents the direction of the angular acceleration of the bicycle’s front wheel while it is slowing to rest from Mary’s point of view?

$\textbf{(A) }$ Into the ground$ \qquad$ $\textbf{(B) }$ North$ \qquad$ $\textbf{(C) }$ South$ \qquad$ $\textbf{(D) }$ East$ \qquad$ $\textbf{(E) }$ West

$\textbf{E}$

Looking at the front wheel as it travels to the South, the wheel spins forward. By the right-hand rule, this means that the direction of the angular velocity points to the left which would be Eastward. Now, since the angular speed is decreasing, this means that the angular acceleration is in the direction opposite to the angular velocity. This means that the angular acceleration is directed to the West.

For the inclined plane shown, a force of 150 N applied parallel to the incline’s surface is needed to pull the object of mass $M=10.0\ \text{kg}$ along the entire length of the incline’s surface at a constant speed. What is the actual mechanical advantage (AMA) for this scenario?

$\textbf{A}$

The actual mechanical advantage is computed as resistive force divided by applied force. This gives $\text{AMA}=\dfrac{F_{resistive}}{F_{applied}}=\dfrac{100\ \text{N}}{150\ \text{N}}=\dfrac23$.

Two blocks sit on a horizontal frictionless surface connected to an ideal spring. Initially everything is at rest and there is a string compressing the mass-spring system from equilibrium. At some time after the string is cut, the block of mass $2M$ reaches its maximum kinetic energy $K$. What maximum kinetic energy does the block of mass $M$ attain in terms of $K$?

$\textbf{(A) }$ $\dfrac14K \qquad$ $\textbf{(B) }$ $\dfrac12K \qquad$ $\textbf{(C) }$ $K \qquad$ $\textbf{(D) }$ $2K \qquad$ $\textbf{(E) }$ $4K$

$\textbf{D}$

Linear momentum of the system is conserved during the oscillation of the masses. By calling the speed of the heavy mass $v$ when the kinetic energy is maximized for the $2M$ mass, this means by linear momentum conservation that the lighter mass $M$ will have speed $2v$. The kinetic energy of the $2M$ mass is $K=\dfrac12(2M)v^2=Mv^2$ and for the light mass we have $K^\prime=\dfrac12M(2v)^2=2Mv^2=2K$.

For the circuit shown, the four light bulbs have identical resistance, all wires have zero resistance, and the battery is assumed to be ideal with emf $\xi$. When the switch, S, in the circuit is closed, a wire of zero resistance is added into the circuit. Which of the light bulbs will be dimmer after the switch is closed?

An object with mass $M$ moves due East on a frictionless horizontal surface with a speed of $V$. A second object of mass $\dfrac12M$ has a speed of $3V$. The two objects collide and stick together. If the objects are moving due South after the collision, with what speed are they moving?

$\textbf{(A) }$ $\dfrac53V \qquad$ $\textbf{(B) }$ $\dfrac13V \qquad$ $\textbf{(C) }$ $\dfrac{\sqrt5}{3}V \qquad$ $\textbf{(D) }$ $\dfrac{4\sqrt2}{3}V \qquad$ $\textbf{(E) }$ $\dfrac{\sqrt{33}}{3}V$

$\textbf{C}$

The East-West component of linear momentum canceled each other, hence we know the second object has a velocity component of $2V$ to the west. The remaining component of velocity to the south, by Pythagorean Theorem, is $\sqrt{(3V)^2-(2V)^2}=\sqrt5V$. Thus the whole system has a linear momentum of $p=\left(\dfrac12M\right)(\sqrt5V)=\dfrac{\sqrt5}{2}MV$ to the south. The two objects stick together after collision, so the final speed is $v=\dfrac{p}{M+1/2M}=\dfrac{\sqrt5}{3}V$.

A 2.0 kg particle travels at a constant speed of 5.0 m/s along the line shown in the figure. What is the magnitude of the particle’s angular momentum calculated from the origin?

$\textbf{B}$

As shown in the diagram above, the red line $r_\perp$ is perpendicular to the direction of velocity. We can find that $r_\perp=3\sin\theta=3\left(\dfrac45\right)=\dfrac{12}{5}$. Thus the angular $L=r_\perp p=mvr_\perp=(2)(5)\left(\dfrac{12}{5}\right)=24\ \text{kg}\cdot\text{m}^2\text{/s}$.

$\textbf{A}$

The conventional current is directed counterclockwise in the circuit, meaning from a right-hand rule that the magnetic field interior to the loop is directed out of the plane of the page. As a result, the magnetic field at P (outside the loop) would be oppositely directed to the interior or into the page’s plane. Since the resistance is decreasing, the current in the outer circuit increases, thereby increasing the field strength through the inner circuit. By Lenz’s Law, there is an induction to fight the change in field and there is a current in the inner circuit producing a magnetic field into the page. This means that the current is directed clockwise and hence, from X to Y in the inner circuit’s resistor.

A parallel-plate capacitor of capacitance $C$ has an insulating dielectric material of constant $\kappa$ filling the entire region between its plates. This capacitor now is connected to a battery of voltage $V$ and is fully charged before the battery is disconnected. The capacitor stores energy $U$. Using insulating gloves, a person now removes the dielectric from the capacitor. After equilibrium is established following the removal of the dielectric, which one of the following choices best represents the energy stored by the capacitor in terms of $U$?

$\textbf{(A) }$ $\kappa^2U \qquad$ $\textbf{(B) }$ $\kappa U \qquad$ $\textbf{(C) }$ $U \qquad$ $\textbf{(D) }$ $U/\kappa \qquad$ $\textbf{(E) }$ $U/\kappa^2$

$\textbf{B}$

By disconnecting the battery, the charge on the capacitor plates is fixed. By removing the dielectric, the capacitance is reduced by a factor of $\kappa$. So, by using $U=\dfrac12\dfrac{Q^2}{C}$, with the reduction in the capacitance by $\kappa$ with no charge change, the energy is increased by a factor of $\kappa$.

A standing wave is created in an air-filled tube open at both ends when a tuning fork of frequency 552 Hz is placed near one of the open ends. A tuning fork of frequency 644 Hz produces the next highest harmonic for this tube. What is the length of the open tube? Assume room temperature.

$\textbf{(A) }$ 0.46 m$ \qquad$ $\textbf{(B) }$ 0.92 m$ \qquad$ $\textbf{(C) }$ 1.59 m$ \qquad$ $\textbf{(D) }$ 1.85 m$ \qquad$ $\textbf{(E) }$ 3.70 m

$\textbf{D}$

Since consecutive standing wave modes for tubes open at both ends have a difference in frequency equal to the fundamental frequency, we know that the fundamental frequency of the tube is $644-552=92\ \text{Hz}$. So, for the 1st harmonic, we can write $\lambda=\dfrac{v}{f}=\dfrac{340}{92}=3.70\ \text{m}$. For this standing wave mode, there is only 1/2 of a wavelength within the tube of length $L$. Hence, the tube length is $L=\dfrac12\lambda=1.85\ \text{m}$.

One mole of a monatomic ideal gas undergoes an isobaric expansion. In the process, the temperature of the gas increases from 300 K to 500 K. Which one of the following choices best represents the amount of work done by the gas on the surroundings?

$\textbf{(A) }$ 665 J$ \qquad$ $\textbf{(B) }$ 1110 J$ \qquad$ $\textbf{(C) }$ 1660 J$ \qquad$ $\textbf{(D) }$ 2490 J$ \qquad$ $\textbf{(E) }$ 4160 J

$\textbf{C}$

The ideal gas equation $PV=nRT$ can be rewrote as $P\Delta V=nR\Delta T$ when the pressure $P$ is a constant. Thus the work done by the gas to the surroundings is $W=P\Delta V=nR\Delta T=(1)(8.31)(500-300)=1662\ \text{J}$.

An object is placed in front of a thin converging lens resulting in a focused image forming on a screen. If the bottom half of the lens were covered with black paper, what happens to the image on the screen?

$\textbf{(A) }$ There is no change of any kind to the image.$ \qquad\newline$
$\textbf{(B) }$ Only the top half of the object is focused into an image on the screen.$ \qquad\newline$
$\textbf{(C) }$ Only the top half of the object is focused into an image on the screen and it now is half the size.$ \qquad\newline$
$\textbf{(D) }$ The image is fully formed, only it now is half as large.$ \qquad\newline$
$\textbf{(E) }$ The image is fully formed, only it is dimmer.

$\textbf{E}$

Even with half of the lens covered, there is still a full image formed, but only half as many rays from the object are focused onto the screen. This means that a full image will form but it will be less intense and hence, will appear dimmer.

A hollow conducting sphere in static equilibrium is isolated in deep space with a net excess charge $+Q$ on it. What is the electric potential (assuming $V(r\rightarrow\infty)=0$) at the position labeled P shown in the interior of the figure? The sphere has radius $R=3a$ and the point of interest is at a location $r=2a$ from the center of the sphere at an angle of $60^\circ$ with respect to the $+x-\text{axis}$.

$\textbf{C}$

By symmetry, the charge $+Q$ is equally distributed on the surface, thus the potential of the sphere is $\dfrac{kQ}{3a}$. Since there is no charge inside of the sphere, no electric lines exist inside of the sphere either. So the electric potential in the interior of the shell is the same as the surface.

A simple pendulum of length $L$ has a point mass $M$ released from rest from the horizontal position shown. In the absence of air resistance and friction, the mass swings through the arc of a circle. Let $T$ represent the magnitude of the force from the string on the mass (tension), $G$ represent the magnitude of the gravitational force acting on the mass by the earth, and $F$ represent the magnitude of the net force acting on the mass. Which one of the following choices describes the relationship among these forces when the mass swings at the bottom of the arc (point P)?

$\textbf{A}$

The gravitational force $G=mg$. From M to P, the potential energy is converted into kinetic energy, thus we have $\dfrac12mv_P^2=mgL$. So the speed of mass $M$ at point P is $v_P=\sqrt{2gL}$. The net force of the circular motion at point P is $F=T-G=m\dfrac{v_P^2}{L}=2mg$. So $T=F+G=2mg+mg=3mg$. Finally we get $G<F<T$.

An electromagnetic wave has a magnetic field given by the expression (in Cartesian coordinates) $\vec{B}(x,y,z,t)=\left(6.0\times10^{-6}\right)\cos(2.21\times10^7z-6.63\times10^{15}t)\hat{x}$. At time $t=0$ and position $x=y=z=0$, what is the direction of the electric field associated with this wave?

$\textbf{(A) }$ $+x \qquad$ $\textbf{(B) }$ $-x \qquad$ $\textbf{(C) }$ $+y \qquad$ $\textbf{(D) }$ $-y \qquad$ $\textbf{(E) }$ $+z$

$\textbf{D}$

The EM wave associated with this magnetic field travels along the $+z$ direction from the argument in the cosine term. The magnetic field at the position and time given points along the $+x$ direction since $\cos(0)=1$. The Poynting vector gives the direction of energy flow for the EM wave which is also the same as the direction of travel of the wave. Hence, writing that $\vec{S}~\vec{E}\times\vec{B}$, we need to figure out $\vec{?}\times\hat{x}=\hat{z}$. By Right-hand rules $-\hat{y}\times\hat{x}=\hat{z}$, and so the electric field is directed along $-y$ here.

An inverted “V” in static equilibrium is made from two uniform beams each of mass $M=12\ \text{kg}$. Each beam of the “V” has the same length and makes an angle of $30^\circ$ with the vertical as shown in the diagram. Which one of the following choices best represents the magnitude of the static friction force acting on the left leg of the “V” from the level ground? The coefficient of static friction between each beam and the ground is $\mu_s=0.76$.

$\textbf{B}$

From the free body diagram, the contacting force $C$ by the right beam is equal to the frictional force $f$. Performing a torque analysis with the ground touching point as the pivot, we have $Mg\cdot\dfrac{L}2\sin30^\circ=CL\cos30^\circ$, which gives $C=f=\dfrac12Mg\tan30^\circ=\dfrac12(12)(10)\dfrac{\sqrt3}3=34.6\ \text{N}$.

A person sets a one-meter long stick so that it makes a $30^\circ$ angle with the $x-\text{axis}$. An observer in a space ship moving along the $x-\text{axis}$ measures the angle of the stick to be $60^\circ$ with the $x-\text{axis}$. With what speed is the space ship moving in terms of the speed of light, $c$?

$\textbf{(A) }$ $\dfrac12c \qquad$ $\textbf{(B) }$ $\dfrac{\sqrt2}3c \qquad$ $\textbf{(C) }$ $\dfrac34c \qquad$ $\textbf{(D) }$ $\dfrac2{\sqrt5}c \qquad$ $\textbf{(E) }$ $\dfrac{2\sqrt2}{3}c$

$\textbf{E}$

For the person sitting on the stick, the $x$ component of the length is $x=(1)\cos30^\circ=\dfrac{\sqrt3}2\ \text{m}$ and the $y$ component is $y=(1)\sin30^\circ=\dfrac12\ \text{m}$. From the view of the observe in the space ship, the $x$ component of length is $x^\prime=x\sqrt{1-\dfrac{v^2}{c^2}}$ and the $y$ component $y^\prime=y$, so he measures the angle $\tan60^\circ=\dfrac{y^\prime}{x^\prime}=\dfrac{\frac12}{\frac{\sqrt3}{2}\sqrt{1-v^2/c^2}}=\sqrt3$. So the speed of the space ship is $v=\sqrt{\dfrac89}c=\dfrac{2\sqrt2}{3}c$.